Question

A solid cubical block of wood has dimensions as shown in the figure (3ft side), and the block is to be cut in half as indicated by the shaded region.
Column A: The total surface area of one of the resulting halves of the block
Column B: 36 square feet

• A. Quantity A is greater
• B. Quantity B is greater
• C. The two quantities are equal
• D. The relationship cannot be determined from the information given

Hint:

When an object is cut, the total surface area of the resulting pieces is always greater than the original surface area because new surfaces are created. The total surface area of one half will be half the original surface area PLUS half the area of the new cuts.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires calculating the surface area of one of two identical wedges created by cutting a cube along a diagonal plane.
Step 2: Detailed Explanation:
The original block is a cube with side length \(s = 3\) ft.
When the cube is cut in half diagonally (from the top-front edge to the bottom-back edge, for example), each half is a triangular prism (or wedge). The surface of this wedge consists of some of the original faces of the cube plus the new face created by the cut.
The surface of one half consists of 5 faces: 1. Two triangular faces: These are the original front and back faces, each cut in half. They are right triangles with legs of length 3 ft. The area of one such triangle is \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5\) sq ft. Since there are two, their combined area is \(2 \times 4.5 = 9\) sq ft. 2. Two square faces: These are two of the original, uncut faces of the cube (e.g., the bottom and one side face). The area of each is \(s^2 = 3^2 = 9\) sq ft. Their combined area is \(2 \times 9 = 18\) sq ft. 3. One rectangular face (the cut surface): This is the new surface created by the cut. Its width is one edge of the cube (\(s=3\)). Its length is the diagonal of a square face. The length of this diagonal is \(\sqrt{s^2 + s^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\) ft. The area of this rectangle is \(3 \times 3\sqrt{2} = 9\sqrt{2}\) sq ft.
The total surface area of one half is the sum of the areas of these faces:
\[ \text{Total Area} = (\text{Area of 2 triangles}) + (\text{Area of 2 squares}) + (\text{Area of cut rectangle}) \] \[ \text{Total Area} = 9 + 18 + 9\sqrt{2} = 27 + 9\sqrt{2} \text{ sq ft} \] Step 3: Comparing the Quantities:
We need to compare Column A (\(27 + 9\sqrt{2}\)) with Column B (36).
Subtract 27 from both quantities: - New Column A: \(9\sqrt{2}\) - New Column B: \(36 - 27 = 9\)
Divide both by 9: - Newest Column A: \(\sqrt{2}\) - Newest Column B: 1
Since \(\sqrt{2}\) is approximately 1.414, we know that \(\sqrt{2} \textgreater 1\).
Therefore, the quantity in Column A is greater.